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# Permutation and Combination

HI guys: I am trying to put the permutation and combination in my head but still have problem to figure it out if the problem is about P or C. I have a question and the answer of that question is P(9,5). I just want to know how i know its a permutation or combination by just looking at the problem. I understand that permutation means order matters and combination means order doesnt.

There are nine empty seats in a theater, and five customers need to find places to sit. How many different ways can these five seat themselves ?
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mustish1
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Commented:
there are 9 seats for 1st person, 8 seats for 2nd(because 1st person is alreasy in one), 7 seats for 3rd etc.
so permutation (9,5) is 9x8x7x6x5.

Combination would be like asking how many pairs of people can you make from those 5. which would be (5x4)/(2x1)
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Commented:
"different ways"  - that means you care about ordering.

therefore it's a permutation
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Commented:
Combinations would be more like:

There are nine empty seats in a theater, and five basketballs you need to store in them.
How many different ways can you arrange the basketballs ?

This is slightly tricky.  People are different, so order matters.  Basketballs are the same so it doesn't.
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Author Commented:
Thanks d-glitch. I google the topic and try to find something for practice as i still have problem to figure it out if its a P or C

http://www.regentsprep.org/Regents/math/algtrig/ATS5/PCPrac.htm

i found 1 but im still looking more to understand the difference between them
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Commented:
>>  This is slightly tricky.  People are different, so order matters.  Basketballs are the same so it doesn't.

That is not correct.  People are different, and the seats are different too.  That's why it is a permutation.

Contrast:  You have nine people and five small umbrellas (no sharing).  How many arrangements are there.
This is back to combinations.
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Commented:
>understand the difference between them

Try with small numbers, higher than 6 items and the numbers get so large you get lost in the fog.

Take three items, a, b and c.

I can make three pairs from these items, ab, ac and bc <- those are three combinations.

If the order matters, such as how this pair sit in a theatre then I have six different arrangements because I can reverse the pairs:ab, ba, ac, ca, bc and cb <-these are permutations and there are more of them because I don't only find the number of groups, but I find how many ways these groups can be arranged within themselves.

Just as a start at taking three from four items, there are four groups (abc, abd, acd, bcd) <- combinations.
To find number of permutations I also need to multiply by the number of arrangements of each of the three unique groups. for each group there are 6 perms (abc, acb, bac, bca, cab, cba) etc.
So picking three from four gives four combinations, but 6x4 permutations.
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Author Commented:
Thanks RobinD
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